Difference between revisions of "1998 AIME Problems/Problem 14"

(Solution)
Line 31: Line 31:
 
<cmath>(m' - 2) + (n' - 2) \leq (m' - 2)(n' - 2) + 1 = 9 + d.</cmath>
 
<cmath>(m' - 2) + (n' - 2) \leq (m' - 2)(n' - 2) + 1 = 9 + d.</cmath>
 
So
 
So
<cmath>d(8 + 4(m + n)) < 8  + 4(9 + d)</cmath>
+
<cmath>d(8 + 4(m + n)) < 8  + 4(9 + d)</cmath> //shouldn't this be 24 + 4(9+d)? then the entire proof is fallacious (it results in 2(1 + 3 + 3) < 15 which is true)
 
<cmath>d(4 + 4(m + n)) < 44</cmath>
 
<cmath>d(4 + 4(m + n)) < 44</cmath>
 
<cmath>d(1 + m + n) < 11</cmath>
 
<cmath>d(1 + m + n) < 11</cmath>

Revision as of 18:04, 7 July 2014

Problem

An $m\times n\times p$ rectangular box has half the volume of an $(m + 2)\times(n + 2)\times(p + 2)$ rectangular box, where $m, n,$ and $p$ are integers, and $m\le n\le p.$ What is the largest possible value of $p$?

Solution

$2mnp = (m+2)(n+2)(p+2)$

Let’s solve for $p$:

$(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)$

$[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)$

$p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4}$

For the denominator, we will use a factoring trick (colloquially known as SFFT), which states that $xy + ax + ay + a^2 = (x+a)(y+a)$.

$p = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}$

Clearly, we want to minimize the denominator, so $(m-2)(n-2) - 8 = 1 \Longrightarrow (m-2)(n-2) = 9$. The possible pairs of factors of $9$ are $(1,9)(3,3)$. These give $m = 3, n = 11$ and $m = 5, n = 5$ respectively. Substituting into the numerator, we see that the first pair gives $130$, while the second pair gives $98$. We can quickly test for the denominator assuming other values to convince ourselves that $1$ is the best possible value for the denominator. Hence, the solution is $130$.


Proof that setting the denominator $(m - 2)(n - 2) - 8$ to $1$ is optimal: Suppose $(m - 2)(n - 2) = 9$, and suppose for the sake of contradiction that there exist $m', n'$ such that $(m' - 2)(n' - 2) = 8 + d$ for some $d > 1$ and such that \[\frac{2(m + 2)(n + 2)}{(m - 2)(n - 2) - 8} < \frac{2(m' + 2)(n' + 2)}{(m' - 2)(n' - 2) - 8}.\] This implies that \[d(m + 2)(n + 2) < (m' + 2)(n' + 2),\] and \[d((m - 2)(n - 2) + 4(m + n)) < (m' - 2)(n' - 2) + 4(m' + n').\] Substituting gives \[d(9 + 4(m + n)) < 8 + d + 4(m' + n'),\] which we rewrite as \[d(8 + 4(m + n)) < 24 + 4((m' - 2) + (n' - 2)).\] Next, note that for $p'$ to be positive, we must have $m' - 2$ and $n' - 2$ be positive, so \[(m' - 2) + (n' - 2) \leq (m' - 2)(n' - 2) + 1 = 9 + d.\] So \[d(8 + 4(m + n)) < 8  + 4(9 + d)\] //shouldn't this be 24 + 4(9+d)? then the entire proof is fallacious (it results in 2(1 + 3 + 3) < 15 which is true) \[d(4 + 4(m + n)) < 44\] \[d(1 + m + n) < 11\] Next, we must have that $m - 2$ and $n - 2$ are positive, so $3 \leq m$ and $3 \leq n$. Also, $2 \leq d$ by how we defined $d$. So \[2(1 + 3 + 3) < 11,\] a contradiction. We already showed above that there are some values of $m$ and $n$ such that $(m - 2)(n - 2) = 9$ that work, so this proves that one of these pairs of values of $m$ and $n$ must yield the maximal value of $p$.

See also

1998 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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